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Jordan triple product∗
Algeboy†
2013-03-21 21:34:52
Given a Jordan algebra J we define a ternary product on J by
{xyz} = (x.y).z + (y.z).x − (z.x).y.
When J is a special Jordan algebra of characteristic not 2, we know the
product x.y = 21 (xy + yx). In this context a computation shows
{xyz} =
1
(xyz + zyx).
2
This gives a simple method by which to compute the triple product in a special
Jordan algebra.
A key instance of the Jordan triple product is the case when x = z (setting
x.2 = x.x for notation). Here we get
{xyx} = 2(x.y).x − x.2 .y.
In a special Jordan algebra this becomes {xyx} = xyx in the associative product.
To treat a Jordan algebra as a quadratic Jordan algebra this product plays the
role of one of the two quadratic operations: Uy : y 7→ {xyx}. The other is the
usual squaring product y 7→ y .2 .
To establish uniform proof for Jordan algebra in characteristic 2 and also
include exceptional Jordan algebras it is often preferable to encode computations
using the unary product: x.2 and the triple product {xyz}. The connection
between the triple product and the quadratic unary product is found in the
Jordan identity:
(a.2 .b).a = a.2 .(b.a)
This idea was exploited by McCrimmon to establish quadratic Jordan algebras
and many uniform and previously unknown results on Jordan algebras.
The triple product can be compared to the Jacobi product on an algebra
with a [, ] multiplication
#xyz# = [[x, y], z] + [[y, z], x] + [[z, x], y].
∗ hJordanTripleProducti
created: h2013-03-21i by: hAlgeboyi version: h38615i Privacy
setting: h1i hDefinitioni h17C05i
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To make a closer parallel use the typical assumption that [z, x] = −[x, z] (outside
of characteristic 0) then we can write:
#xyz# = [[x, y], z] + [[y, z], x] − [[x, z], y].
Since Jordan algebras are commutative we can also write
{xyz} = (x.y).z + (y.z).x − (x.z).y.
However, unlike Lie algebras where #xyz# = 0, in Jordan algebras the triple
product is almost never 0.
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